Which of the following numbers is a multiple of 2? ${57,83,96,101,113}$
Solution: The multiples of $2$ are $2$ $4$ $6$ $8$ ..... In general, any number that leaves no remainder when divided by $2$ is considered a multiple of $2$ We can start by dividing each of our answer choices by $2$ $57 \div 2 = 28\text{ R }1$ $83 \div 2 = 41\text{ R }1$ $96 \div 2 = 48$ $101 \div 2 = 50\text{ R }1$ $113 \div 2 = 56\text{ R }1$ The only answer choice that leaves no remainder after the division is $96$ $ 48$ $2$ $96$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $2$ are contained within the prime factors of $96$ $96 = 2\times2\times2\times2\times2\times3 2 = 2$ Therefore the only multiple of $2$ out of our choices is $96$. We can say that $96$ is divisible by $2$.